# Dual vector space - Lagrange Interpolating Polynomial

1. Feb 21, 2013

### SqueeSpleen

I think I solved it a week ago, but I didn't write down all the things and I want to be sure of doing the things right, plus the excersise of writing it here in latex helps me a loot (I wrote about 3 threads and didn't submited it because writing it here clarified me enough to find the answer for my self xD). So feel free to correct me.

Lets be $\alpha _{0},...,\alpha _{n} \in K, \alpha _{i} \neq \alpha _{j}$ if $i\neq j$

To every $i$, $0 \leq i \leq n$, we define $\varepsilon _{\alpha _i}:K_{n}[X]\rightarrow K$ as $\varepsilon _{\alpha _i}(P)=P(\alpha _i)$.

i) Prove that $B = \left \{ \epsilon _{\alpha _{0}},...,\epsilon _{\alpha _{n}} \right \}$ is a basis of $(K_{n}[X])^{*}$

ii) Lets be $B = \left \{ P_{0},...,P_{n} \right \}$ the basis of $K_{n}[X]$ such that $B^{*}=B_{1}$. Prove that the polynomial

$P = \sum _{i=0}^{n} \beta _i . P_i$

is the only polynomial in K[X] of grade minor or equal to n such that,

$\forall i, 0 \leq i \leq n, P(\alpha _{i}) = \beta _{i}$

This polynomial is named the Lagrange's Interpolating Polynomial (irrelevant :P).

iii) Prove that there're real numbers $\left \a_{0},...,a_{n}{ \right \}$ such that, to all $P \in \mathbb{R}_{n}[X]$
$\int _{0}^{1}P(x)dx=\sum _{i=0}^{n}a_{i}.P(\alpha _{i})$
Find $a_{0}, a_{1}$ and $a_{2}$ when $n = 2, a_{0}=1, a_{1}=\frac{1}{2}$ and $a_{2}=0$

Relevant proposition:
Lets be V a K-vector space of dimensión n, and lets be V* it's dual space.
Lets be $B_{1}=\left \{ \varphi _{1},...\varphi _{n} \right \}$ a basis of V*. Then there's only one basis $B = \left \{ v _{1},...v _{n} \right \}$ of V such that:
$B^{*}=B_{1}$

i) Suppose that:
$\sum _{i=0}^{n}\alpha _{i}\varepsilon _{i}=0$
Then, to all $P \in K_n[X]$, we have: $\sum _{i=0}^{n}\alpha _{i}\varepsilon _{i}(P)=\sum _{i=0}^{n}P(\alpha _{i})=0$
To every $\forall i, 0 \leq i \leq n$
We take the polynomial:
$\prod _{i=0,{i\neq }j}^{n}(X-\alpha _{i})$
And evaluating it we know $\alpha _{j}=0$, because $P(\alpha _{i})=0$ to every i≠j $(P(\alpha _{j})$ can't be zero because $\alpha _{j} \neq \alpha _{i}$ if i≠j
Then, we have the linear independency of the set, and as n+1 elements, then it's generated space has dimention n+1 and it's a subset of $(K_{n}[X])*$, so it's a basis of $(K_{n}[X])*$ (because it also has dimension n+1).
ii) So we know $B_{1}$ is a basis of $K_n[X]$, then for the proposition we know that there's only one basis
$B= \left \{ P_{0},...,P_{n} \right \}$ such that $B^{*}=B_{1}$
So, by the definition of dual basis, $\epsilon _{\alpha _{i}}(P_{j})=P_{j}(\epsilon _{\alpha _{0}})=$
1 if i=j
0 if i≠j
Then $P(\epsilon _{\alpha _{j}})=\sum _{i=0}^{n} \beta _i . P_i(\epsilon _{\alpha _{j}})=\beta _{i}$

iii)
I think I'm stuck in 3, but I'll think later after sleeping, I'm probably failing due to being so sleepy.






Last edited: Feb 21, 2013
2. Feb 22, 2013

### SqueeSpleen

I didn't find the edit button in my first post, I assume someone blocked it so I bump the thread when I finish it instead of buring it in the bottom of the forum's abyss.
If $P=\sum _{i=0}^{n}\beta _{i} P_{i}$ then
$\int _{0}^{1}\sum _{i=0}^{n} \beta _{i} . P_{i} (x)dx = \sum _{i=0}^{n} \beta _{i}.\int _{0}^{1}P_{i} (x)dx$
$\sum _{i=0}^{n}a_i.P(\alpha _{i})=\sum _{i=0}^{n}a_i.\sum _{j=0}^{n} \beta _{i} P_{i} (\alpha _{i})=$
$\sum _{i=0}^{n}a_{i}.\beta _{i}$
Let's be $a_{i}=\int _{0}^{1}P_{i} (x)dx$ to all $0 \leq i \leq n$
Then
$\sum _{i=0}^{n} \frac{\beta _{i}}{i+1}=\sum _{i=0}^{n}a_{i}.\beta _{i}$

Let's do find $a_{0}, a_{1}$ and $a_{2}$ when $n = 2, a_{0}=1, a_{1}=\frac{1}{2}$ and $a_{2}=0$
$\left\{\begin{matrix} P_{0}(\alpha _{0})=1\\ P_{0}(\alpha _{1})=0\\ P_{0}(\alpha _{2})=0 \end{matrix}\right.$
$\left\{\begin{matrix} P_{0}(1)=1\\ P_{0}(\frac{1}{2})=0\\ P_{0}(0)=0 \end{matrix}\right.$
As we know two roots of $P_{0}$ we only need to multiply it for a constant to make it equal to 1 when evaluated in 1.
$-\frac{(X-\frac{1}{2})(X)}{2}$

$\left\{\begin{matrix} P_{1}(1)=0\\ P_{1}(\frac{1}{2})=1\\ P_{1}(0)=0 \end{matrix}\right.$
$-4(X-1)(X)$

$\left\{\begin{matrix} P_{2}(1)=0\\ P_{2}(\frac{1}{2})=0\\ P_{2}(0)=1 \end{matrix}\right.$

$2(X-\frac{1}{2})(X-1)$

So we only need to evaluate the integrals of these polynomials.
Then:
$a_{0}=-\frac{1}{24}$
$a_{1}=\frac{2}{3}$
$a_{2}=\frac{1}{6}$

PD: The excersise is from http://mate.dm.uba.ar/~jeronimo/algebra_lineal/AlgebraLineal.pdf page 115 Ejercise 16, this isn't the book used in my Linear Algebra's course in the dual vector space subject but I find the ejercises were pretty interesting to do.

Last edited: Feb 22, 2013